Question

我需要帮助弄清楚如何返回从键盘输入的2d数组中所有数字的总和，这些数字可被3整除。我有我的计算和方法，它将正确返回一个测试用例，但是对于给定的每个案例它都没有正确计算。任何建议都会非常有帮助。我将加载我的代码，包括while循环，以便更容易找到我计算错误的位置。

try
    {
        InputStreamReader stream = new InputStreamReader (System.in);
        BufferedReader scan = new BufferedReader(stream);

        inputParser = Integer.parseInt(input);

        int i = 0;
        while(inputParser != 0)
        {
                input = scan.readLine();
                inputParser = Integer.parseInt(input);
                if(inputParser == 0)
                {
                    inputParser = 0;
                }
                else
                {
                    numbers1[i] = inputParser;
                    i++;
                }
        }

        sum = computeSumDivisibleBy3(numbers1,0,numbers1.length-1);
        System.out.println("The sum of the numbers divisible by 3 is " + sum);

    }
    catch(NumberFormatException exception)
    {
        System.out.println("Please enter integers only");
    }

这里是计算可被3整除的和的方法

//而不是这个原始方法，我已经在你的下方实现了它，它正确返回

public static int computeSumDivisibleBy3(int[] numbers, int startIndex, int endIndex){
    if (startIndex == endIndex)
        return numbers[endIndex];
    else{
        int sum1 = computeSumDivisibleBy3(numbers, startIndex, endIndex-1);
        if (numbers[endIndex] % 3 == 0)
            return sum1 + numbers[endIndex];

        else

    return sum1;

    }

}
//newly implemented code
public static int computeSumDivisibleBy3(int[] numbers, int startIndex, int endIndex){
    if (startIndex == numbers.length-1)
        return numbers[startIndex] % 3 == 0 ? numbers[startIndex] : 0;
    else{

        return (numbers[startIndex] % 3 == 0 ? numbers[startIndex] : 0) + computeSumDivisibleBy3( numbers, ++startIndex, endIndex );

    }

}

Answer 1

// is startIndex really needed? you only ever use it unnecessarily, so, no.
public static int computeSumDivisibleBy3(int[] numbers, int startIndex, int endIndex){
    // End condition, good, but the return value doesn't make any sense
    // You always return the last one in the array, even it's not divisible by 3
    //if (startIndex == endIndex)
    //    return numbers[endIndex];

    // How about this instead (taking the removal of startIndex into consideration)
    if( index == numbers.length - 1) {
        return (numbers[index] % 3 == 0 ? numbers[index] : 0);
    }

    // now, on to regular operations
    // all that's needed is to return the current element, or zero, plus the recursive result
    return (numbers[index] % 3 == 0 ? numbers[index] : 0) + computeSumDivisibleBy3( numbers, ++index );

    // or, if you'd prefer an if statement
    int cur = 0;
    if( current element % 3 is 0 ) {
        cur = current element
    }
    return cur + recurse( numbers, ++index );

    //else{
    //    int sum1 = computeSumDivisibleBy3(numbers, startIndex, endIndex-1);
    //    if (numbers[endIndex] % 3 == 0)
    //        return sum1 + numbers[endIndex];

    //    else

    //return sum1;

    }

}

使用递归计算可被3整除的和

1 个答案: